Problem: Simplify the following expression and state the condition under which the simplification is valid. $z = \dfrac{-9a^2 + 9a + 108}{-7a^3 + 63a^2 - 140a}$
Answer: First factor out the greatest common factors in the numerator and in the denominator. $ z = \dfrac {-9(a^2 - a - 12)} {-7a(a^2 - 9a + 20)} $ $ z = \dfrac{9}{7a} \cdot \dfrac{a^2 - a - 12}{a^2 - 9a + 20} $ Next factor the numerator and denominator. $ z = \dfrac{9}{7a} \cdot \dfrac{(a - 4)(a + 3)}{(a - 4)(a - 5)}$ Assuming $a \neq 4$ , we can cancel the $a - 4$ $ z = \dfrac{9}{7a} \cdot \dfrac{a + 3}{a - 5}$ Therefore: $ z = \dfrac{ 9(a + 3)}{ 7a(a - 5)}$, $a \neq 4$